This is an answer in the special case that $m$ equals $n$.  I have a vague memory that this might be an exercise in the textbook "Algebraic Geometry, A First Course" by Joe Harris.  Let $k$ be an algebraically closed field.

<B>Genus Formula.</B>  For a connected, proper, nodal $k$-curve $C$ with $r$ irreducible components of geometric genus $g_i$ and with $\delta$ nodes, the arithmetic genus of $C$ equals $$p_a(C) = \delta+1-r + \sum_i g_i.$$

Thus, for a complete intersection curve $\text{Zero}(F,G)\subset \mathbb{P}^3_k$ of type $(m,n)$ that is nodal, the maximum possible number of nodes is $$\delta \leq \frac{mn(m+n-2)}{2}.$$

Of course there are complete intersection curves that attain this maximum number.  Let $M_1,\dots,M_m$ and $N_1,\dots,N_n$ be global sections of $\mathcal{O}(1)$.  Let $M$ denote the product $M_1\cdots M_m$.  Let $N$ denote the product $N_1\cdots N_n$.  The zero schemes of $M$ and $N$ are singular hypersurfaces of degree $m$ and $n$: each is a union of hyperplanes.  

<B>Notation.</B>  An ordered tuple of elements of $H^0(\mathbb{P}^3,\mathcal{O}(1))$ is in <b>linearly general position</b> if every subset of $\leq 4$ elements of the tuple is a linearly independent subset of $H^0(\mathbb{P}^3,\mathcal{O}(1))$.  

<B>Lemma.</B>  If $(M_1,\dots,M_m,N_1,\dots,N_n)$ is in linearly general position, then the common intersection $C=\text{Zero}(M,N)$ has $mn(m+n-2)/2$ nodes.  

<B>Proof.</B>  By hypothesis, for every point $p$ in $C$, at least one $M_i$ vanishes at $p$, at least one $N_j$ vanishes at $p$, and at most $3$ of $M_i$ and $N_j$ combined vanish at $p$.  If only one $M_i$ and only one $N_j$ vanish at $p$, then since these linear forms are linearly independent, then they extend to a basis for $H^0(\mathbb{P}^3,\mathcal{O}(1))$, i.e., a system of homogeneous coordinates.  It is straightforward to compute that, Zariski locally, $C$ equals a line near $p$.  Thus, $p$ is a smooth point of $C$.

Thus, assume that $3$ of the $M_i$ and $N_j$ vanish, say $M_1$, $M_2$, and $N_1$.  Again, this extends to a homogeneous coordinate system, so that locally $C$ is isomorphic to a union of two lines in $\mathbb{P}^3$ that intersect at $p$, i.e., $p$ is a node of $C$.  The number of such nodes of type $(M_i,M_{i'},N_j)$ equals $mn(m-1)/2$.  The number of such nodes of type $(M_i,N_j,N_{j'})$ equals $mn(n-1)/2$.  In total, the number of nodes equals $mn(m+n-2)/2$.  <B>QED</B>

<B>Proposition.</B>  Assume that the characteristic of $k$ equals $0$.  With notation as above, if $m=\ell=n$ and $(M_1,\dots,M_\ell,N_1,\dots,N_\ell)$ is in linearly general position, then for a general pair $F,G\in H^0(\mathbb{P}^3,I_C(\ell))$, both $\text{Zero}(f)$ and $\text{Zero}(G)$ are smooth. Also $\text{Zero}(F,G)$ equals $C$.  Thus, there exist complete intersection curves attaining the maximum number of nodes that are equal to a complete intersection of two smooth hypersurfaces.

<B>Proof.</B>  This is a standard application of Bertini's Theorem.  Denote the blowing up of $\mathbb{P}^3$ along the ideal sheaf $I_C$ of $C$ by $$\nu:\widetilde{\mathbb{P}^3}_k \to \mathbb{P}^3_k.$$  For each node $p$ of $C$, there is a unique ordinary double point $\widetilde{p}$ of $\widetilde{\mathbb{P}^3}_k$ that maps to $p$.  The only singularities of $\widetilde{\mathbb{P}^3}_k$ are these finitely many ordinary double points.

The strict transform of the linear system $\text{Span}(M,N)$ defines a projective, surjective morphism from the blowing up to a smooth genus $0$ curve, $$\pi:\widetilde{\mathbb{P}^3}_k \to \Pi.$$  The restriction of $\nu$ to each geometric fiber of $\pi$ is a closed immersion equal to the corresponding divisor in the linear system of divisors on $\mathbb{P}^3_k$.  Since the characteristic is $0$, Generic Smoothness implies that there are only finitely many singular fibers of $\pi$ (necessarily containing the finitely many nodes $\widetilde{p}$).  Thus, for two general fibers, say $\text{Zero}(F)$ and $\text{Zero}(G)$, the corresponding divisors in $\mathbb{P}^3$ are smooth.  By Bezout's Theorem, the intersection $\text{Zero}(F,G)$ is a curve of degree $\ell \cdot \ell$ that contains $C$.  However, already $C$  is a curve of degree $\ell\cdot \ell$.  Thus $\text{Zero}(F,G)$ equals $C$. <B>QED</B>